Write t = n(T/4) + t0 with t0 less than T/4 and φ = ωt − nπ/2. Then s = a[n + 1 − cos φ] when n is even, and s = a[n + sin φ] when n is odd. For your sliders t = s gives n = , φ = degrees and s = cm.
Problem statement
A particle moves along the x axis according to the law x = a cos ωt. Find the distance that the particle covers during the time interval from t = 0 to t.
We assume the motion is simple harmonic and undamped. This means a and ω stay fixed for all time.
We assume "distance" means the total path length, not the displacement. This means the trip back counts as well as the trip out.
We assume t is a general time, written as t = n(T/4) + t0, where n is a whole number and t0 is less than T/4. This means we count whole quarter periods first.
Symbols: a is the amplitude, T = 2π/ω is the period, n is the number of whole quarter periods, and φ = ωt − nπ/2 is the leftover phase. The slider shows t as a share of T, so t/T runs from 0 to 4.
The big idea in plain words
The point goes from one end to the middle and then to the other end, over and over. Each quarter of a period it covers exactly a, the amplitude. So after n whole quarters the distance is n a, no matter where we are in the cycle. Only the leftover part needs care, and it depends on whether the leftover starts at an end (slow) or at the middle (fast).
The solution in each frame
Lab frame
The axis is fixed to the lab, with x = 0 at the equilibrium position. The full solution is in the next section.
Point rest frame
This frame does not apply. The point keeps accelerating, so a frame fixed to it is not inertial, and it does not help to find the path length.
The solution step by step
The point starts at the extreme x = a when t = 0. It reaches the middle at T/4, the opposite extreme at T/2, the middle again at 3T/4 and the start at T. Each of these steps covers the distance a. So we write the time as t = n(T/4) + t0, with t0 less than T/4.
Case 1, n even. After an even number of quarters the point is back at an extreme. The leftover starts there, so we can use x = a cos φ. The point has moved from a down to a cos φ, which is a distance a − a cos φ.
Case 2, n odd. After an odd number of quarters the point is at the middle. The leftover starts there, so the position measured from the middle is a sin φ, which is the distance moved.
For your sliders: n = , so the full legs add up to n a = cm. The leftover is cm, and the total is s = cm.
The book result is the two-case formula above, and the page uses it. A second, independent formula (the area under the speed graph, written as 2k + 1 − cos w) gives the same distance, and a numeric integral of |vx| over the window agrees: cm against cm. The PDF writes one line as "x = x a sin ωt", which is a typo for x = a sin ωt. A finite-difference slope of x gives cm/s against the formula cm/s.
Five methods
1. Force method
The force F = −mω²x gives the law x = a cos ωt, which we used. It does not give distances directly, so it only supplies the motion.
2. Energy method
Energy gives the speed at each place, so it gives the leftover distance only if we know the time. It does not shorten the route. It does tell us the speed is zero at the ends, which is why the point turns round there.
3. Graph method
The distance is the area under the speed against time graph. Each quarter period has the same area a, so the distance graph climbs by a at each dotted line. The leftover is a part of one more hump.
4. Calculus method
This is the step-by-step route: s = ∫ |vx| dt with |vx| = aω |sin ωt|. Over each quarter the integral is a, and the leftover is 1 − cos φ or sin φ times a.
5. Check method
Units: a is a length, so s is a length. Limits: at t = T/4 the formula gives s = a, at t = T it gives 4a, as it must. For small t the formula gives s ≈ a φ²/2, so the point creeps at first, because it starts at rest. The distance never decreases, and it is never less than the displacement.
Fastest: the graph method. Most reliable: the calculus method, with the numeric check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Amplitude a | Yes, proportional | Every distance, the track, the tape, every graph height |
| Period T | Not the distance for the same t/T | The value of t in seconds, the speed sizes, the time axis and the motion speed |
| Time share t/T | Yes | t, n, the leftover phase, the leftover distance, the total s and the tape |
| Screen speed | No | Only how fast the animation runs compared with real time |
| Target distance (given-answer mode) | Sets the answer | Solves t from s by bisection |
| Zoom, Auto-fit | No | Pixels per centimetre in the picture only |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Amplitude | a | cm |
| Period | T | s |
| Time as a share of T | t/T | |
| Time | t | s |
| Whole quarter periods | n | |
| Leftover phase | φ | degrees |
| Distance | s | cm |
Exam traps
- Using x(t) − x(0) as the distance. That is the displacement, which can even be zero after a full period.
- Using the wrong leftover for n odd. After an odd number of quarters the point is at the middle, so the leftover is a sin φ, not a(1 − cos φ).
- Counting n from half periods instead of quarter periods. A quarter period is T/4, not T/2.
- Forgetting that the phase φ = ωt − nπ/2 is the leftover only, not the whole phase ωt.
- Putting the angle in degrees into a formula with ω in rad/s. Keep radians.
- Mixing ω (rad/s) with the frequency f (Hz). ω = 2π/T.